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Let $f: \mathbb R\to \mathbb R$ be defined by $f(x)=|x-1|$. Show that $f$ is neither one one nor onto function.

My Attempt: $$f(x)=|x-1|$$

For all $x \in \mathbb R$, the set of values of $f(x)$ are non negative real numbers. So, range of $f(\mathbb R)=[0, \infty)$. Hence, $f$ is not onto.

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    $\begingroup$ Certainly the not one to one part is clear... what inputs give, say, 1? $\endgroup$ – Sean Roberson Jul 16 '17 at 7:27
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    $\begingroup$ Use $f(2)=f(0)$ $\endgroup$ – Fakemistake Jul 16 '17 at 7:27
  • $\begingroup$ We have $f(x)\geq 0$. $\endgroup$ – Wuestenfux Jul 16 '17 at 8:08
  • $\begingroup$ What is the preimage of, say, ...$-1$? $\endgroup$ – Alvin Lepik Jul 16 '17 at 8:32
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In order to be one to one, it should be valid that for every $x\neq y$ we have that $f(x)\neq f(y)$. Well, try $x=a+1$ and $y=1-a$, for every $a>0$. We have that: $$f(a+1)=|a+1-1|=|a|=|-a|=|1-a-1|=f(1-a)$$ So, $f$ is clearly not one to one.

Since $$f(x)=|x-1|\geq0$$ it is also ivedent that $f$ is not onto $\mathbb{R}$.

Note: If we define $f:\mathbb{R}\to[0,+\infty)$ $f$ would be onto $[0,+\infty)$.

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$f$ is not one-to-one because exists $x\neq y\in\mathbb{R}$ such that $f(x)=f(y)$, for example, $x=0, y=2$.

$f$ is not onto because $f(x)=|x|\ge0$ for all $x\in\mathbb{R}$, then there is no $x$ such that $f(x)=-1$ (for example).

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